Combustion analysis of 1.0g of a compound containing carbon , hydrogen and oxygen gives 2.3 g of carbon dioxide and 0.93g of water . If The relative molecular mass is 58. Calculate The empirical and molecular formula

Where are you with this one?

Convert mass CO2 to mass C.
Convert mass H2O to mass H(atoms).
1.0g - mass C - mass H = mass O

Convert g C to mols.
Convert g H to mols
Convert g O to mols.

Find the ratio for the empirical formula.
Then (emprical formula) x number = molar mass. Solve for number (rounded to whole number).
Then (empirical formula)n = molecular formula.

Good

1.23

To calculate the empirical formula and molecular formula of a compound, we need to follow a step-by-step process using the given information.

Step 1: Convert the masses of carbon dioxide (CO2) and water (H2O) to moles. This can be done by dividing the given masses by their respective molar masses.

The molar mass of CO2 = 12.01 g/mol (C) + (2 * 16.00 g/mol) (O) = 44.01 g/mol
The moles of CO2 = 2.3 g / 44.01 g/mol ≈ 0.052 moles

The molar mass of H2O = (2 * 1.01 g/mol) (H) + 16.00 g/mol (O) = 18.02 g/mol
The moles of H2O = 0.93 g / 18.02 g/mol ≈ 0.052 moles

Step 2: Determine the number of moles of carbon, hydrogen, and oxygen in the compound. Since the ratio of moles in a compound gives us the empirical formula, we need to find the smallest whole number ratio.

In CO2, there is 1 mole of carbon and 2 moles of oxygen.
In H2O, there are 2 moles of hydrogen and 1 mole of oxygen.

Step 3: Calculate the number of moles of carbon, hydrogen, and oxygen per mole of compound by multiplying the moles obtained in step 2 by their respective subscripts in the molecular formula.

For carbon, the number of moles is 1 * 0.052 moles = 0.052 moles.
For hydrogen, the number of moles is 2 * 0.052 moles = 0.104 moles.
For oxygen, the number of moles is 2 * 0.052 moles = 0.104 moles.

Step 4: Divide the number of moles of each element by the smallest number of moles obtained in step 3 to get the simplest whole number ratio.

Carbon: 0.052 moles / 0.052 moles = 1
Hydrogen: 0.104 moles / 0.052 moles = 2
Oxygen: 0.104 moles / 0.052 moles = 2

Therefore, the empirical formula is CH2O.

Step 5: Calculate the empirical formula mass by summing up the molar masses of the elements in the empirical formula.

Empirical formula mass = 12.01 g/mol (C) + (2 * 1.01 g/mol) (H) + 16.00 g/mol (O) = 30.03 g/mol

Step 6: Calculate the empirical formula mass ratio by dividing the molecular mass given in the question by the empirical formula mass.

Empirical formula mass ratio = 58 g/mol / 30.03 g/mol ≈ 1.93

Step 7: Multiply each subscript in the empirical formula by the empirical formula mass ratio obtained in step 6 to get the molecular formula.

Molecular formula = (C1H2O1)1.93 = C1.93H3.86O1.93

However, we cannot have fractional subscripts in a molecular formula, so we need to round them to the nearest whole number.

Molecular formula ≈ C2H4O2

Therefore, the empirical formula is CH2O, and the molecular formula is C2H4O2.